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How to solve a Poisson equation with only Neumann boundaries?

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Old   March 31, 2009, 13:14
Default How to solve a Poisson equation with only Neumann boundaries?
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I need to solve a pressure Poisson equation with only Nuemann boundaries with F.D. method. How we can get the unique results without introducing any errors into?


Many thanks in advance!
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Old   March 31, 2009, 20:46
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Specify a pressure value (any value) at just one point (anywhere) on the boundary.
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Old   April 1, 2009, 03:23
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But if we set up a reference point at the boundary, certain error will introduced. any possibility we do not do that?

Cheers!
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Old   April 1, 2009, 09:56
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> certain error will introduced.

What kind of error? There is no unique exact solution to this problem, not just for discretely but analytically: the solution is unique only up to an additive constant.
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Old   April 1, 2009, 11:12
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yes, that is the thing. If we set one of the point on the boundary as reference point, which is Dirichlet bounday. But how we can make sure it is also satisfying the Nuemann boundary condition? If it is not, error on the boundary will be introduced.

Cheers!
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Old   April 2, 2009, 02:56
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Quote:
Originally Posted by mbgnfrx2 View Post
yes, that is the thing. If we set one of the point on the boundary as reference point, which is Dirichlet bounday. But how we can make sure it is also satisfying the Nuemann boundary condition? If it is not, error on the boundary will be introduced.

Cheers!
You may prescribe the reference pressure value at any point, not necessarily on the boundary.
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Old   April 2, 2009, 04:01
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Originally Posted by mbgnfrx2 View Post
I need to solve a pressure Poisson equation with only Nuemann boundaries with F.D. method. How we can get the unique results without introducing any errors into?


Many thanks in advance!
When you construct your matrix A.
Make Albnd = 0, and do not add its contribution to Ap (or diagonal element).

Imagine that your Ap (diagonal element) is like this

Ap = -Sum (Al)

To get Ap you usually do Ap = Ap - Albnd
Src = Src + albnd * phi (if phi was fixed value).

Instead now you do not modify Src. Do not modify Ap, just make Albnd = 0.

solve matrix.

You get some solution, pick a point from these Phi you just calculated. Lets say iref.

phi_ref = Phi[ iref ] ;

For all
{
Phi = Phi_from_matrix - phi_ref.
}


Thats all. Painless and simple.
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